A 60 mm jet leaves a nozzle at 36 m/s and is deflected through 120°. Compute the force on the blade when it is stationary, then when the blade moves away at 5.4 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.060)^{2}/4 = 0.00283 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00283)(36)^{2}(1 - \cos 120^{\circ}) = 5,497 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 36 - 5.4 = 30.6 m/s$$

6. Substituting

   $$F = 1000(0.00283)(30.6)^{2}(1 - \cos 120^{\circ}) = 3,971 N$$

7. Power

   $$P = F u = 3,971(5.4) = 21.44 kW$$

Why the other options are there

• 824.5 N (mixed absolute and relative velocity)
• P = 29.68 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 40 mm jet leaves a nozzle at 38 m/s and is deflected through 90°. Compute the force on the blade when it is stationary, then when the blade moves away at 15.2 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.040)^{2}/4 = 0.00126 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00126)(38)^{2}(1 - \cos 90^{\circ}) = 1,815 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 38 - 15.2 = 22.8 m/s$$

6. Substituting

   $$F = 1000(0.00126)(22.8)^{2}(1 - \cos 90^{\circ}) = 653.3 N$$

7. Power

   $$P = F u = 653.3(15.2) = 9.93 kW$$

Why the other options are there

• 725.8 N (mixed absolute and relative velocity)
• P = 27.58 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 30 mm jet leaves a nozzle at 43 m/s and is deflected through 120°. Compute the force on the blade when it is stationary, then when the blade moves away at 21.5 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.030)^{2}/4 = 0.00071 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00071)(43)^{2}(1 - \cos 120^{\circ}) = 1,960 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 43 - 21.5 = 21.5 m/s$$

6. Substituting

   $$F = 1000(0.00071)(21.5)^{2}(1 - \cos 120^{\circ}) = 490.1 N$$

7. Power

   $$P = F u = 490.1(21.5) = 10.54 kW$$

Why the other options are there

• 980.2 N (mixed absolute and relative velocity)
• P = 42.15 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 55 mm jet leaves a nozzle at 30 m/s and is deflected through 180°. Compute the force on the blade when it is stationary, then when the blade moves away at 7.5 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.055)^{2}/4 = 0.00238 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00238)(30)^{2}(1 - \cos 180^{\circ}) = 4,276 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 30 - 7.5 = 22.5 m/s$$

6. Substituting

   $$F = 1000(0.00238)(22.5)^{2}(1 - \cos 180^{\circ}) = 2,406 N$$

7. Power

   $$P = F u = 2,406(7.5) = 18.04 kW$$

Why the other options are there

• 1,069 N (mixed absolute and relative velocity)
• P = 32.07 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 45 mm jet leaves a nozzle at 29 m/s and is deflected through 60°. Compute the force on the blade when it is stationary, then when the blade moves away at 8.7 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.045)^{2}/4 = 0.00159 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00159)(29)^{2}(1 - \cos 60^{\circ}) = 668.8 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 29 - 8.7 = 20.3 m/s$$

6. Substituting

   $$F = 1000(0.00159)(20.3)^{2}(1 - \cos 60^{\circ}) = 327.7 N$$

7. Power

   $$P = F u = 327.7(8.7) = 2.85 kW$$

Why the other options are there

• 200.6 N (mixed absolute and relative velocity)
• P = 5.82 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 75 mm jet leaves a nozzle at 20 m/s and is deflected through 180°. Compute the force on the blade when it is stationary, then when the blade moves away at 7.0 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.075)^{2}/4 = 0.00442 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00442)(20)^{2}(1 - \cos 180^{\circ}) = 3,534 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 20 - 7.0 = 13.0 m/s$$

6. Substituting

   $$F = 1000(0.00442)(13.0)^{2}(1 - \cos 180^{\circ}) = 1,493 N$$

7. Power

   $$P = F u = 1,493(7.0) = 10.45 kW$$

Why the other options are there

• 1,237 N (mixed absolute and relative velocity)
• P = 24.74 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 35 mm jet leaves a nozzle at 20 m/s and is deflected through 60°. Compute the force on the blade when it is stationary, then when the blade moves away at 10.0 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.035)^{2}/4 = 0.00096 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00096)(20)^{2}(1 - \cos 60^{\circ}) = 192.4 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 20 - 10.0 = 10.0 m/s$$

6. Substituting

   $$F = 1000(0.00096)(10.0)^{2}(1 - \cos 60^{\circ}) = 48 N$$

7. Power

   $$P = F u = 48(10.0) = 0.48 kW$$

Why the other options are there

• 96 N (mixed absolute and relative velocity)
• P = 1.92 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 65 mm jet leaves a nozzle at 32 m/s and is deflected through 180°. Compute the force on the blade when it is stationary, then when the blade moves away at 11.2 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.065)^{2}/4 = 0.00332 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00332)(32)^{2}(1 - \cos 180^{\circ}) = 6,796 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 32 - 11.2 = 20.8 m/s$$

6. Substituting

   $$F = 1000(0.00332)(20.8)^{2}(1 - \cos 180^{\circ}) = 2,871 N$$

7. Power

   $$P = F u = 2,871(11.2) = 32.16 kW$$

Why the other options are there

• 2,379 N (mixed absolute and relative velocity)
• P = 76.11 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 40 mm jet leaves a nozzle at 37 m/s and is deflected through 180°. Compute the force on the blade when it is stationary, then when the blade moves away at 5.6 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.040)^{2}/4 = 0.00126 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00126)(37)^{2}(1 - \cos 180^{\circ}) = 3,441 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 37 - 5.6 = 31.5 m/s$$

6. Substituting

   $$F = 1000(0.00126)(31.5)^{2}(1 - \cos 180^{\circ}) = 2,486 N$$

7. Power

   $$P = F u = 2,486(5.6) = 13.80 kW$$

Why the other options are there

• 516.1 N (mixed absolute and relative velocity)
• P = 19.10 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade

A 60 mm jet leaves a nozzle at 20 m/s and is deflected through 120°. Compute the force on the blade when it is stationary, then when the blade moves away at 2.0 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.060)^{2}/4 = 0.00283 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00283)(20)^{2}(1 - \cos 120^{\circ}) = 1,696 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 20 - 2.0 = 18.0 m/s$$

6. Substituting

   $$F = 1000(0.00283)(18.0)^{2}(1 - \cos 120^{\circ}) = 1,374 N$$

7. Power

   $$P = F u = 1,374(2.0) = 2.75 kW$$

Why the other options are there

• 169.6 N (mixed absolute and relative velocity)
• P = 3.39 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Moving Blade